4MA0 Higher June 2017 Paper 3R Q11
11 There are 9 red counters and 11 blue counters in a bag.
There are no other counters in the bag.
Emeka takes at random a counter from the bag and writes down the colour of the counter.
He puts the counter back in the bag.
Natasha takes at random a counter from the bag and writes down the colour of the counter.

| Scheme | Marks |
|---|---|
Complete correct binary structure for selection of two counters OR At least one additional red branch labelled \(\dfrac{9}{20}\) and at least two blue branches labelled \(\dfrac{11}{20}\) | M1 |
| Correct probability tree diagram | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\dfrac{9}{20} \times \dfrac{11}{20}\) | M1 |
| \(\dfrac{99}{400}\) | A1 |
| (2) |
Notes
M1: for \(\text{‘}\dfrac{9}{20}\text{’} \times \text{‘}\dfrac{11}{20}\text{’}\)
A1: (ft diagram) for \(\dfrac{99}{400}\) or 0.2475 or 24.75%
| Scheme | Marks |
|---|---|
| \(\text{‘}\dfrac{9}{20}\text{’} \times \text{‘}\dfrac{9}{20}\text{’}\) or \(\text{‘}\dfrac{11}{20}\text{’} \times \text{‘}\dfrac{11}{20}\text{’}\) | M1 |
| \(\text{‘}\dfrac{9}{20}\text{’} \times \text{‘}\dfrac{9}{20}\text{’} + \text{‘}\dfrac{11}{20}\text{’} \times \text{‘}\dfrac{11}{20}\text{’}\) | M1 |
| \(\dfrac{202}{400}\) | A1 |
| (3) | |
| (7 marks) |
Notes
M1: for one correct product
ft from diagram
M1: for the complete method
ft from diagram
A1: \(\dfrac{202}{400}\) oe or 0.505 or 50.5%
Alternative method
M1 for \(2 \times \text{‘}\dfrac{9}{20}\text{’} \times \text{‘}\dfrac{11}{20}\text{’}\) oe (ft from (a))
M1 for \(1 - 2 \times \text{‘}\dfrac{9}{20}\text{’} \times \text{‘}\dfrac{11}{20}\text{’}\)